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A coil of inductance $L = 50 μ H$ and resistance $R = 0.5 Ω$ is connected to a battery of emf = 5V. A resistance of $10 Ω$ is connected parallel to the coil. Now, at some instant the connection of the battery is switched off. Then, the amount of heat generated in the coil after switching off the battery is

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1.25 mJ

2.5 mJ

0.65 mJ

0.12 mJ

answer is D.

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Detailed Solution

$U = \frac{1}{2} L i_{0}^{2} = \frac{1}{2} \times 50 \times 10^{- 6} {(\frac{5}{0.5})}^{2} J = 2.5 m J$

Now, after switching off the battery this energy is dissipated in coil (resistance = $0.5 Ω$ ) and resistance ( $10 Ω$ ) in the ratio of their resistances ( $H = i^{2} R t$ or $H \propto R$ ). Therefore, heat generated in the coil.

$\begin{array}{rcl} H & = & 2.5 (\frac{0.5}{0.5 + 10}) m J \\ = & 0.12 m J \end{array}$

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